पाटीगणित (श्रीधराचार्य - प्राचीन संस्कृत टीका एवं प्रो. कृपाशंकर शुक्ल सविस्तार व्याख्या)

MIXTURES. SPECIAL PROBLEMS 63
Since the visible quantity stands near a square root, therefore per-
forming the operations prescribed in the first part of the rule, we get
[ (√(4 × 8 + 2²) + 2) / 2 ]², i.e., 16.
This is the number which when diminished by 2 times the square root
of itself yields 8 as remainder.
Now treat this (16) as the visible quantity. Since it stands near a
fraction, therefore performing the operation prescribed in the second part
of the rule, we get
16 / (1 - 1/5), i.e., 20.
This is the number which when diminished by 1/5 of itself yields 16
as remainder.
Now treat this (20) as the visible quantity. Since it stands near a
square root, therefore performing the operations prescribed in the first
part of the rule, we get
[ (√(4 × 20 + 1) + 1) / 2 ]², i.e., 25.
This is the number which when diminished by its square root yields 20 as
remainder.
Now treat this (25) as the visible quantity. Since this stands near a
fraction, therefore performing the operation prescribed in the second part
of the rule, we get
25 / (1 - ⅙), i.e., 30.
This is the number which when diminished by its ⅙ yields 25 as
remainder.
Now treat this (30) as the visible quantity. Since this stands near a
square root, therefore performing the operations prescribed in the first
part of the rule, we get
[ (√(4 × 30 + 1) + 1) / 2 ]², i.e., 36.
This is the number which being diminished by its square root yields 30
as remainder.
Since all quantities in the statement of the problem are now
exhausted, no further operation is needed. The required number is thus
obtained to be 36.